If the area of the region enclosed by the cur | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The area $A$ of the region enclosed by the curve $x^2+y^2=16$ and the lines $x=2$ and $x=3$ is given by the integral:$A = \int_2^3 \sqrt{16-x^2} d

Vedclass · Accepted Answer

$16 \sin^{-1}\left(\frac{3}{4}\right)$

Vedclass · Answer

(A) The area $A$ of the region enclosed by the curve $x^2+y^2=16$ and the lines $x=2$ and $x=3$ is given by the integral:$A = \int_2^3 \sqrt{16-x^2} dx$Using the standard integral formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}ight)$:$A = \left[\frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}\left(\frac{x}{4}ight)ight]_2^3$Evaluating the definite integral:$A = \left(\frac{3}{2}\sqrt{16-9} + 8\sin^{-1}\left(\frac{3}{4}ight)ight) - \left(\frac{2}{2}\sqrt{16-4} + 8\sin^{-1}\left(\frac{2}{4}ight)ight)$$A = \left(\frac{3}{2}\sqrt{7} + 8\sin^{-1}\left(\frac{3}{4}ight)ight) - \left(\sqrt{12} + 8\sin^{-1}\left(\frac{1}{2}ight)ight)$Since $\sqrt{12} = 2\sqrt{3}$ and $\sin^{-1}\left(\frac{1}{2}ight) = \frac{\pi}{6}$:$A = \frac{3}{2}\sqrt{7} + 8\sin^{-1}\left(\frac{3}{4}ight) - 2\sqrt{3} - 8\left(\frac{\pi}{6}ight)$$A = \frac{3}{2}\sqrt{7} - 2\sqrt{3} - \frac{4\pi}{3} + 8\sin^{-1}\left(\frac{3}{4}ight)$Comparing this with the given expression $\left(3 \sqrt{7}-4 \sqrt{3}-\frac{8 \pi}{3}+kight)$:Note that the given expression is $2 	imes \left(\frac{3}{2}\sqrt{7} - 2\sqrt{3} - \frac{4\pi}{3} + \frac{k}{2}ight)$.Actually,evaluating the integral result directly: $A = \frac{3}{2}\sqrt{7} - 2\sqrt{3} - \frac{4\pi}{3} + 8\sin^{-1}\left(\frac{3}{4}ight)$.To match the form $\left(3 \sqrt{7}-4 \sqrt{3}-\frac{8 \pi}{3}+kight)$,we multiply the expression by $2$ and divide by $2$:$A = \frac{1}{2} \left(3\sqrt{7} - 4\sqrt{3} - \frac{8\pi}{3} + 16\sin^{-1}\left(\frac{3}{4}ight)ight)$Thus,$k = 16\sin^{-1}\left(\frac{3}{4}ight)$.

If the area of the region enclosed by the curve $x^2+y^2=16$ and the lines $x=2$ and $x=3$ is $\left(3 \sqrt{7}-4 \sqrt{3}-\frac{8 \pi}{3}+k\right)$ sq units,then $k$ equals

Similar Questions

Let $y$ be the function which passes through $(1, 2)$ having slope $(2x + 1)$. The area bounded between the curve and $x$-axis is

The area of the region bounded by the curve $y=2x-x^2$ and the $x$-axis is

The area of the region bounded by the curve $y = \cos x$,$x = \frac{\pi}{2}$,$x = \frac{3\pi}{2}$,and $y = 0$ is . . . . . . sq. units.

Find the area under the given curve $y=x^{2}$ bounded by the lines $x=1$,$x=2$ and the $x$-axis.

The area enclosed by the ellipse $25x^2 + 16y^2 = 400$ is . . . . . . sq. units. (in $\pi$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module